Answer:
(a)Time taken to reach the ground, x=6 seconds
(b)Maximum height of the arrow during its flight =101.25 feet
Step-by-step explanation:
Given the quadratic function which models the flight of the arrow:
[tex]y=-5x^2+15x+90[/tex]
Where:
- y represents the vertical height of the arrow, in meters; and
- x represents the time, in seconds.
(a)Positive Zero
[tex]-5x^2+15x+90=0\\-5(x^2-3x-18)=0\\-5(x^2-6x+3x-18)=0\\-5(x(x-6)+3(x-6))=0\\-5(x+3)(x-6)=0\\x+3=0\:or\:x-6=0\\x=-3,x=6[/tex]
In 6 seconds, the arrow will hit the horizontal plane, i.e. the ground.
(b)
[tex]y=-5x^2+15x+90\\y-90=-5(x^2-3x)\\y-90-5(\frac{9}{4}) =-5(x^2-3x+(\frac{3}{2} )^2)\\y-101.25=-5(x-\frac{3}{2} )^2\\y=-5(x-\frac{3}{2} )^2+101.25[/tex]
Comparing with the vertex form:
[tex]y=a(x-h)^2+k[/tex]
Our vertex, (h,k)=(1.5, 101.25)
Maximum Value of y=101.25
The maximum value of y represents the maximum height of the arrow during its flight, which is 101.25 feet.
